If antpool goes and tries a 51% attack then it's no longer 14%, it's a larger % since you have 28EH/s going against what is now only 172EH/s
It's 14% of the total hashrate though.
A 51% attack is defined as 51% of the
total hashrate, which includes both honest and dishonest mining. If there is a total of 100 EH/s, a 51% attacker would control 51 EH/s, leaving 49 EH/s in the hands of the honest miners. If the 51% was based against the hashrate of the honest miners, then an attacker with 34 EH/s would have 51% of the remaining honest 66 EH/s, but obviously 34 EH/s isn't enough to beat 66 EH/s
Correct me if I've miscalculated.
You have.
Since
p and
q are probabilities, they always sum to 1.
p is the probability an honest miner finds the next block. When there is no attack,
p = 1.
q is the probability an attacker finds the next block.
Let's say a pool with 14% of the hashrate turns evil. The remaining 86% of the hashrate stays honest.
p = 0.86 and
q = 0.14. That's simple enough.
Let's say, on the other hand, 100% of the current hashrate stays honest, but an outside miner with the equivalent of 14% of this hashrate starts attacking the network. Call it 100 EH/s honest hashrate and an outside miner comes along with 14 EH/s dishonest hashrate. There is now a total of 114 EH/s. The attacker with 14 EH/s has a 14/114 = 12.3% chance of finding the next block. So, in this scenario,
p = 0.877 and
q = 0.123.
Greg makes it clear on the tool you are using - the number you are entering is the "Proportion of hash-power", not the ratio of dishonest to honest hash power.